9.10 Going Deeper: Enrichment & Exam Preparation

Where the Theorem Shows Up in Real Life

Math in the Real World

  • The builder’s 3-4-5 rule. Masons and carpenters set out a perfect right angle without any instrument by knotting a rope into lengths 3, 4 and 5. Pulled taut into a triangle, the corner between the 3-side and the 4-side is exactly 90^\circ, because 3^2 + 4^2 = 5^2. The same trick squares up foundations, tiled floors and cricket pitches.
  • Screen and TV sizes. A screen’s advertised size is its diagonal. A monitor 24 cm wide and 18 cm tall has a diagonal of \sqrt{24^2 + 18^2} = \sqrt{900} = 30 cm, that is just the (3,4,5) triple scaled by 6. Knowing the theorem, you can recover the diagonal from the width and height of any rectangular display.
  • Shortest distance and navigation. A ship sails 12 km east then 5 km north; its straight-line distance from port is \sqrt{12^2 + 5^2} = 13 km, far shorter than the 17 km it actually travelled. GPS units and map apps use exactly this “horizontal step and vertical step” idea to measure straight-line separations.
  • Ramps, ladders and cable runs. Wheelchair ramps, leaning ladders and tent guy-ropes are all hypotenuses of right triangles. A ladder 15 m long with its foot 9 m from a wall reaches \sqrt{15^2 - 9^2} = 12 m up, the safe height is a side calculation, not a guess.
  • A bridge from earlier and later chapters. The theorem runs entirely on the squares and square roots of Chapter 1, and reappears whenever we measure diagonals, heights and distances in later geometry and coordinate work.

Quick Reference: Triples and Formulas

Memorising a handful of triples turns many problems into instant recognition, spot two sides of a known triple and the third drops out without any arithmetic.

Primitive triple (a,b,c) Check a^2+b^2=c^2 From m,n
(3,4,5) 9+16=25 m=2,\,n=1
(5,12,13) 25+144=169 m=3,\,n=2
(8,15,17) 64+225=289 m=4,\,n=1
(7,24,25) 49+576=625 m=4,\,n=3
(20,21,29) 400+441=841 m=5,\,n=2
(9,40,41) 81+1600=1681 m=5,\,n=4
(12,35,37) 144+1225=1369 m=6,\,n=1
(11,60,61) 121+3600=3721 m=6,\,n=5

Every other triple is a multiple of a primitive one. Scaling (3,4,5) alone gives an endless family:

k 1 2 3 4 5 10
(3k,4k,5k) (3,4,5) (6,8,10) (9,12,15) (12,16,20) (15,20,25) (30,40,50)

The triple-generating formula. For any integers m > n > 0, a = m^2 - n^2,\qquad b = 2mn,\qquad c = m^2 + n^2 gives a triple; it is primitive when m,n share no common factor and are not both odd.

Use the grid-distance explorer below to see the theorem turn two points into a single distance:

Distance between two grid points

One-Page Revision

Quick Revision Card

  • The theorem: in a right triangle, a^2 + b^2 = c^2, where c is the hypotenuse (opposite the right angle) and always the longest side.
  • Find the hypotenuse: c = \sqrt{a^2 + b^2}.  Find a leg: b = \sqrt{c^2 - a^2} (subtract, because you already know the longest side).
  • Isosceles right triangle: c = a\sqrt{2}, so the diagonal of a square of side a is a\sqrt{2}.
  • Triples to know by heart: (3,4,5),\ (5,12,13),\ (8,15,17),\ (7,24,25),\ (20,21,29),\ (9,40,41), and all their multiples.
  • Scaling rule: if (a,b,c) works, so does (ka,kb,kc) for every whole number k.
  • Generating formula: a=m^2-n^2,\ b=2mn,\ c=m^2+n^2 reaches every primitive triple.
  • Grid distance: \sqrt{(\text{horizontal step})^2 + (\text{vertical step})^2}.
  • \sqrt{2}\approx 1.414 is irrational, never a fraction, never a terminating decimal.

Spot the Mistake

Common Exam Mistakes

  • Adding when you should subtract. To find a leg you compute \sqrt{c^2 - a^2}, not \sqrt{c^2 + a^2}. The hypotenuse is the biggest number, so it sits alone on one side.
  • Putting the hypotenuse in the wrong place. c is always opposite the right angle. Squaring the two short sides and adding only works when c is truly the longest side.
  • Calling any close trio a triple. (6,7,9) looks tidy but 6^2 + 7^2 = 85 \neq 81 = 9^2. Always check a^2+b^2=c^2 exactly.
  • Forgetting to scale back. (9,12,15) is a valid triple (it is 3\times(3,4,5)), so do not reject it just because it is not primitive.
  • Writing \sqrt{a^2+b^2}=a+b. Roots do not split over a sum: \sqrt{9+16}=\sqrt{25}=5, not 3+4=7.
  • Losing the units. A length comes out in cm or m, an area in cm², keep track of which the question wants.

Exam Tip Before reaching for a calculator, glance at the two given sides and ask, “Are these part of a triple I know?” Legs 20 and 21? That is the (20,21,29) triple, the hypotenuse is 29 with no arithmetic at all. Sides 9 and 41? Spot (9,40,41) and the missing leg is 40. Recognising scaled triples like (6,8,10) or (15,36,39) saves precious minutes.

Right-Triangle Detective

Enter three side lengths and the tool decides whether they form a right triangle, and, if so, which side is the hypotenuse. Test the triples, then try near-misses like 6, 7, 9.

Is this a right triangle?
Type three side lengths (any order) and press Check.
Tested 0 · Right triangles 0

Exam Corner: CBSE-Style Practice

Mixed Practice (objective, short, long, HOTS)

Objective type (1 mark each)

  1. In a right triangle the hypotenuse is the side opposite the ______ angle.
  2. The hypotenuse of a right triangle with legs 20 and 21 is ______.
  3. Which of (6,7,9), (8,15,17), (4,5,6) is a Pythagorean triple?

Short answer (2 marks each)

  1. A right triangle has hypotenuse 25 cm and one leg 7 cm. Find the other leg.
  2. Find the distance between the grid points (2,3) and (14,8).

Long answer (3 marks each)

  1. A 17 m ladder leans against a vertical wall with its foot 8 m from the base. How high up the wall does it reach? If the foot is then pulled out to 15 m, how high does it reach?
  2. A rectangular park is 40 m long and 9 m wide. A jogger runs along the two sides from one corner to the opposite corner; a cyclist takes the straight diagonal path. How much shorter is the cyclist’s route?

HOTS (Higher Order Thinking)

  1. Using the formula a=m^2-n^2,\ b=2mn,\ c=m^2+n^2, generate the triple for m=5,\ n=2, and verify it satisfies a^2+b^2=c^2. Is it primitive?
  2. A flagpole snaps so that the broken top, still attached at the break, touches the ground 9 m from the base while the standing stump is 12 m tall. What was the original height of the flagpole?

Assertion–Reason (Choose: (a) both true, R explains A; (b) both true, R does not explain A; (c) A true, R false; (d) A false, R true.)

  1. Assertion (A): (20,21,29) is a primitive Pythagorean triple.   Reason (R): A triple is primitive when its three numbers share no common factor greater than 1.
  1. The hypotenuse is opposite the right (90^\circ) angle.
  2. c = \sqrt{20^2 + 21^2} = \sqrt{400 + 441} = \sqrt{841} = \mathbf{29}. (This is the (20,21,29) triple.)
  3. (8,15,17), since 8^2 + 15^2 = 64 + 225 = 289 = 17^2. (For (6,7,9): 36+49 = 85 \neq 81; for (4,5,6): 16+25 = 41 \neq 36.)
  4. Other leg = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = \mathbf{24} cm. (The (7,24,25) triple.)
  5. Horizontal step = 14 - 2 = 12, vertical step = 8 - 3 = 5. Distance = \sqrt{12^2 + 5^2} = \sqrt{169} = \mathbf{13} units.
  6. Foot 8 m: height = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = \mathbf{15} m (the (8,15,17) triple). Foot 15 m: height = \sqrt{17^2 - 15^2} = \sqrt{289 - 225} = \sqrt{64} = \mathbf{8} m. (Pulling the foot out lowers the reach from 15 m to 8 m.)
  7. Diagonal = \sqrt{40^2 + 9^2} = \sqrt{1600 + 81} = \sqrt{1681} = 41 m. The jogger covers 40 + 9 = 49 m, so the cyclist’s route is 49 - 41 = \mathbf{8} m shorter.
  8. a = 5^2 - 2^2 = 21, b = 2(5)(2) = 20, c = 5^2 + 2^2 = 29. Check: 21^2 + 20^2 = 441 + 400 = 841 = 29^2. ✓ Since \gcd(20,21,29) = 1, it is primitive ((20,21,29)).
  9. The standing stump (leg 12) and the ground distance (leg 9) give the broken length as the hypotenuse: \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 m. Original height = stump + broken part = 12 + 15 = \mathbf{27} m.
  10. (a), Both statements are true and R correctly explains A: 20^2 + 21^2 = 29^2, and \gcd(20,21,29) = 1, so the triple is primitive exactly because its terms share no common factor.

Connections to Other Chapters

How This Chapter Links Across the Book

  • Chapter 1 (A Square and a Cube): the whole theorem is built from squares and square roots, a^2, b^2, c^2 and the \sqrt{\ } that recovers a side.
  • Irrational numbers: \sqrt{2} here is the first number proven to be neither a fraction nor a terminating decimal, opening the door to the real-number line.
  • Coordinate geometry and distance: the grid-distance rule \sqrt{(\Delta x)^2 + (\Delta y)^2} is the seed of the distance formula used throughout later coordinate work.
  • Areas of triangles and quadrilaterals: finding the height of an equilateral triangle, the side of a rhombus from its diagonals, or the diagonal of a rectangle all lean on this one theorem.

Glossary

Key Terms

  • Hypotenuse: the side of a right triangle opposite the right angle; always the longest side.
  • Leg (perpendicular side): either of the two shorter sides that meet at the right angle.
  • Baudhayana–Pythagoras Theorem: a^2 + b^2 = c^2 for a right triangle with hypotenuse c.
  • Pythagorean / Baudhayana triple: three positive integers (a,b,c) with a^2 + b^2 = c^2, e.g. (3,4,5).
  • Primitive triple: a triple whose three numbers share no common factor greater than 1.
  • Scaled (non-primitive) triple: a triple obtained by multiplying a primitive one by a whole number k, such as (6,8,10).
  • Irrational number: a number, like \sqrt{2}, that cannot be written as a fraction \tfrac{m}{n} of whole numbers.